Numerical answer to Partial derivative of Hopfield-style Energy in symbolic expression

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Pseudoscientist 2022-7-28

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评论： Torsten 2022-8-2

采纳的回答： Torsten

Hello,

I'm implementing a gradient descent optimizer that uses Hopfield-style energy as function,

the hopfield-style energy is a function of p:

The hopfield energy equation can also be expressed as:

The gradient vector is defined as:

I'm attempting to calculate the gradient vector using the Symbolic Toolbox as follows:

r_mat = sym('r',[4,4]);

r_mat = r_mat(:);

p_mat = sym('p',[4*4,1]);

c_mat = sym('c',[4*4,1]);

Energy = -0.5*p_mat'*r_mat*p_mat' - c_mat'*p_mat;

%the first partial derivative:

q1 = -1*diff(testi1,p_mat(1,1));

which yields as the first term:

conj(c1) + r1_1*conj(p1) + (r1_2*conj(p5))/2 + (r1_3*conj(p9))/2 + (r2_1*conj(p2))/2 + (r1_4*conj(p13))/2 + (r2_2*conj(p6))/2 + (r2_3*conj(p10))/2 + (r3_1*conj(p3))/2 + (r2_4*conj(p14))/2 + (r3_2*conj(p7))/2 + (r3_3*conj(p11))/2 + (r4_1*conj(p4))/2 + (r3_4*conj(p15))/2 + (r4_2*conj(p8))/2 + (r4_3*conj(p12))/2 + (r4_4*conj(p16))/2

How can I replace the symbols with numbers to yield the q vector? I need the q-vector for the gradient descent

the algorithm I'm implementing: https://doi.org/10.1117/1.JEI.23.1.013007 (relevant part starts at Equation 5)

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Answer 1

Torsten 2022-7-28

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1769895-numerical-answer-to-partial-derivative-of-hopfield-style-energy-in-symbolic-expression#answer_1017030

编辑：Torsten 2022-7-28

在 MATLAB Online 中打开

p is (4x1), not (16x1).

And keep r_mat as matrix, then

E = -0.5*p.'*r_mat*p - c.'*p

( r_mat can be assumed symmetric ).

By the way:

dE/dp = -r_mat*p - c

assuming r_mat is symmetric.

To replace symbols by numerical values, use "subs".

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Pseudoscientist 2022-8-2

It turned out symbolic expressions was extremely slow, I used symbolic to calculate the partial derivative and then used that to think how to write the equation in vector form and was evetually succesful and the optimizer is now very fast !